# Tag Info

29

The given sum is what we would get if we were to compute the definite But we can also compute this directly: Now by amazing coincidence $9.3$ squared happens to evaluate to which is good enough to pass for $86.5$. One more lucky coincidence: $9.3$ is easily multiplied by $\frac 23$. So we need to do $6.2$ times $86.5$ which yields The $.3$ cancels ...

25

Revised answer: Grandpa's words are important here - we mustn't move the digits... or in fact do anything with the digits at all. Because what he wants us to do is: See as follows: Initial answer (revised after OP post clarification at +4): The two halves of this equation are equivalent if we write them as: Specifically:

24

Yes: $\sqrt{1}$ is an integer. After that, no. If you've multiplied up to $n$, Bertrand's postulate states that there's always a new prime somewhere from $\frac12n$ to $n$. So there's always at least one prime that doesn't have a partner inside the square root: in other words, the number $1×2×3×\cdots$ cannot be a perfect square.

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There are a few tricksy ways to make this work (e.g. by bending semantics with a birthday on 29 February in Southern Hemisphere summertime, and claiming that 28 years later - when the Gregorian calendar usually repeats itself, with all dates falling on exactly the same days of the week - it's technically "only my seventh birthday", for instance...)....

18

I am officially an idiot. I spent several hours figuring out brilliant deductions and got really good progress with many actual numbers on the grid, and even though I got stuck at places, there was always some clever bit that got me just that much forward. In the end, I was just about to fill the grid in two different ways to show that the puzzle must be ...

18

The only possibilities are: Proof:

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if you select any 85 pets in this shop, then there will invariably be at least one hamster, at least one ferret, at least one chinchilla, and at least one guinea pig

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Yet another way:

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I rotated the grid 90 degrees clockwise before solving the grid deductions, in order to reduce vertical space. Let's solve Statue Park first. Now to the Fillomino. Now that the two grids are solved, we can combine them in two ways: Turns out that the second interpretation is correct, and the column sums are which spells out... As @oAlt noticed, the ...

14

Four circles: Five circles: Six circles: I used mixed integer linear programming to find these optimal solutions, and the first several optimal values are:

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I'm not sure about the bonus but here's the best you can do for the main part Proof that this is the best Lower bound for the bonus Improving this bound for larger $n$ (credit to Greg Martin in the comments for this idea)

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The smallest currency denomination is 1 cent. Percentage needs to be integer. Answer:

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Thanks to Stiv for solving the final cryptic clue You did this to highlight the Matriarch mostly prepared oats for American airline location? (8) AIRLINE + LOCATION = ? 51 7581, 674321 29138 85421'8 141738 (four)

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Aunt Isabel received Reasoning After looking at the numbers Therefore the totals are, by child's age

10

If I understand correctly, then we are looking at solutions to the following equation: It is easy to get all possible values of $N$: This step could be done by hand, e.g. by considering different possibilities of $N \mod 10$. But I just wrote a quick program, as it's more efficient. Thus at most there are expected participants.

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The number chosen is somewhere between $10$ and $99$. So the final answer is

9

Non-programming solution Say there were originally $3N$ people scheduled to attend the event, but only $2N$ actually came. Then so we must solve the equation Now we have This gives us an upper bound. Now a quick calculation (just one calculation!) verifies that this is indeed the maximal possible solution, which means the answer is

9

Assuming we can use the floor function (denoted here with brackets $\lfloor\cdot\rfloor$): So we can make Then we could simply do: and Edited to add: If floor function is allowed we can make each of the following using a single four: and the following using two fours: then we could do:

9

No. Assume the top-left is odd. Then top-right and bottom-right are odd, and bottom-left must be both odd and even, because TL (odd) + BL = 13, so BL is even, but BL - BR (odd) = 6, so BL must also be odd. Therefore the top-left is not odd. However, the same argument applies for TL being even, so this has no solution.

8

1 = 1 2 = 2 3 = 3 4 = 4 5 = 5 6 = 5 + 1 7 = 5 + 2 8 = 5 + 3 9 = 5 + 4 10 = 5 + 4 + 1 11 = 5 + 4 + 2 12 = 5 + 4 + 3 13 = 5 + 4 + 3 + 1 14 = 5 + 4 + 3 + 2 15 = 5 + 4 + 3 + 2 + 1 16 = 4 ⋅ (1 + 3) 17 = (5 ⋅ 4) - 3 18 = (5 ⋅ 4) - 2 19 = (5 ⋅ 4) - 1 20 = (5 ⋅ 4) 21 = (5 ⋅ 4) + 1 22 = (5 ⋅ 4) + 2 23 = (5 ⋅ 4) + 3 24 = (...

8

I finally got it! Below is a detailed explanation

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Another way:

7

I'll take a shot at it Arrived at as follows First Then So But So

6

Because the total score for all three contestants over three events The highest possible $b$ is Since Marina placed second in rhythmic gymnastics, With these values, there is only one way to get Sakura and Hina's scores of 10 and 9 respectively;

6

(I'm understanding the problem to mean that each if the six distances appears exactly once.) Explanation:

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Here is a solution inspired by Beastly Gerbil's observation in a comment that: From here, we can finish with almost no casework: We didn't even use the condition in the third column, though we can check that it's satisfied at the end.

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The rule is: Explicitly: and so Of course, with a question like this there's no way to know for sure that this is the definitive 'right' answer, as you could come up with any number of different rules that give $7 \times 3 = 5$ and $9 \times 5 = 41$. But this "feels'' like it's the intended solution.

6

Bass has shown that there are "obviously" multiple solutions. It appears that there are in fact exactly 24. from pysmt.shortcuts import Symbol, LE, GE, And, Int, Equals, NotEquals, Plus, Minus, Times from pysmt.typing import INT grid = [[Symbol(f"g{row}{col}", INT) for col in range(6)] for row in range(6)] inrange = And(And(GE(grid[row][...

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